# WhaT are 2 consecutive odd numbers that have the product of 12,099?

Jun 27, 2015

I found:
$109 \times 111 = 12099$
or
$- 111 \times - 109 = 12099$

#### Explanation:

Consider the condition for your two consecutive odd numbers:
$\left(2 n + 1\right) \times \left(2 n + 3\right) = 12 , 099$
rearranging:
$4 {n}^{2} + 8 n - 12096 = 0$
Using the Quadratic Formla you should get:
${n}_{1} = - 56$
${n}_{2} = 54$
So you have in $\left(2 n + 1\right) \times \left(2 n + 3\right) = 12 , 099$:
$109 \times 111 = 12099$
or
$- 111 \times - 109 = 12099$

Jun 27, 2015

I got the same answer as Gio (of course), but set it up a little differently.

#### Explanation:

When asked to find odd (or even) numbers, it is often a good idea to use $2 n + 1$ for an odd number (and $2 n$ for an even).

In a problem like this, I like to simply get an equation, find all solution and then pick out the odd (or even) solutions if there are any.

I notice that $12 , 099$ is very close to $12 , 100$, And also recalling that $121 = {11}^{2}$, so $121 \times 100 = {11}^{2} \times {10}^{2}$

Of course consecutive odd numbers are exactly $2$ away from each other, so

let's call our numbers $x - 1$ and $x + 1$

We want $\left(x - 1\right) \left(x + 1\right) = 12 , 099$

So solve ${x}^{2} - 1 = 12 , 099$ (Which we've sort of already done!)

${x}^{2} = 12 , 100$

$x = \pm 110$

So the numbers are:

$110 - 1 = 109$ and $110 + 1 = 111$ (yes, they are odd)

and

$- 110 - 1 = - 111$ and $- 110 + 1 = - 109$ (yes, they're odd too)